Answer
$$\frac{1}{3}\left( {3x + 1} \right)\left[ {\ln \left( {3x + 1} \right) - 1} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \left( {3x + 1} \right)} dx,\,\,\,\,u = 3x + 1 \cr
& {\text{Using the given substitution}} \cr
& u = 3x + 1,\,\,\,\,\,\,\,du = 3dx,\,\,\,\,\,dx = \frac{{du}}{3}\, \cr
& {\text{write in terms of }}u \cr
& \int {\ln \left( {3x + 1} \right)} dx = \int {\ln u} \left( {\frac{{du}}{3}} \right) \cr
& = \frac{1}{3}\int {\ln u} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 11}} \cr
& \left( {11} \right):\,\,\,\,\,\,\int {\ln u} du = u\ln u - u + C \cr
& \frac{1}{3}\int {\ln u} du = \frac{1}{3}u\ln u - \frac{1}{3}u + C \cr
& {\text{write in terms of }}x{\text{, and replace }}3x + 1{\text{ for }}u \cr
& = \frac{1}{3}\left( {3x + 1} \right)\ln \left( {3x + 1} \right) - \frac{1}{3}\left( {3x + 1} \right) + C \cr
& {\text{factoring}} \cr
& = \frac{1}{3}\left( {3x + 1} \right)\left[ {\ln \left( {3x + 1} \right) - 1} \right] + C \cr} $$
